Compute LCM of Two Numbers
“Compute LCM of Two Numbers” is a basic and important programming exercise problem, Here, we are given two numbers, entered by user and our task is to compute LCM (Lowest Common Multiple) of two numbers.
LCM: An LCM of two integers is a smallest possible integer which is divisible by both the numbers.
Example:
INPUT: N1 = 72, N2 = 120 OUTPUT: 360
There are two possible ways to compute the LCM of two numbers.
- Brute-Force, Using loops.
- Using Formula, with GCD.
METHOD 1: Brute-Force
The steps required to find the LCM of two Numbers are as follows:
- Scan the Numbers from user.
- Find the maximum of two number and store it into ‘max’ variable.
- Iterate a loop starting from max until an integer is found which is divisible by both the integers.
- That integer will be the LCM.
C++ Program to compute the LCM of two numbers is as follows:
#include<bits/stdc++.h> using namespace std; int main() { //Scan the number int n1,n2; cout<<"Enter the numbers:"; cin>>n1>>n2; // Computing the LCM int max = (n1 > n2) ? n1 : n2; int LCM; while(1) { if(max % n1 == 0 && max % n2 ==0) { LCM = max; break; } max++; } // Printing the LCM cout<<"The LCM of "<<n1<<" and "<<n2<<" is "<<LCM; }
OUTPUT: Enter the numbers: 72 120 The LCM of 72 and120 is 360
METHOD 2: Using GCD
We have seen in this post: How to find GCD. The LCM and GCD are interrelated, so we can find the LCM of two numbers using GCD of given numbers, as:
LCM = (n1 * n2) / GCD
C++ Program to compute the LCM of two numbers:
#include<bits/stdc++.h> using namespace std; int main() { //Scan the number int n1,n2; cout<<"Enter the numbers:"; cin>>n1>>n2; // Computing the GCD and then LCM int GCD = 1; for (int i = 1; i <= n1 && i <= n2; ++i) { if (n1 % i == 0 && n2 % i == 0) GCD = i; } int LCM = (n1 * n2) / GCD; // Printing the LCM cout<<"The LCM of "<<n1<<" and "<<n2<<" is "<<LCM; }
OUTPUT: Enter the numbers: 72 120 The LCM of 72 and120 is 360
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